Algebra - Simplifying Complex Fractions - Free Printable

Let’s solve each complex fraction one by one. We’ll simplify them step by step, just like cleaning up a messy room — we take it piece by piece.

---

Problem 1:
\[
\frac{\frac{7}{3}}{\frac{28}{9}}
\]

This is a fraction divided by another fraction. To divide fractions, we multiply by the reciprocal (flip the bottom fraction).

So:
\[
\frac{7}{3} \div \frac{28}{9} = \frac{7}{3} \times \frac{9}{28}
\]

Now multiply straight across:

Numerator: \(7 \times 9 = 63\)
Denominator: \(3 \times 28 = 84\)

So we have:
\[
\frac{63}{84}
\]

Now simplify this fraction. Both 63 and 84 are divisible by 21.

\(63 ÷ 21 = 3\)
\(84 ÷ 21 = 4\)

Final answer for Problem 1:
\[
\boxed{\frac{3}{4}}
\]

---

Problem 2:
\[
\frac{\frac{3}{x} + 2}{\frac{5}{x}}
\]

First, let’s combine the terms in the numerator. The numerator is \(\frac{3}{x} + 2\). To add these, write 2 as \(\frac{2x}{x}\):

\[
\frac{3}{x} + \frac{2x}{x} = \frac{3 + 2x}{x}
\]

Now the whole expression becomes:
\[
\frac{\frac{3 + 2x}{x}}{\frac{5}{x}}
\]

Again, dividing fractions → multiply by reciprocal:

\[
\frac{3 + 2x}{x} \times \frac{x}{5}
\]

The \(x\) in the denominator and the \(x\) in the numerator cancel out:

\[
\frac{3 + 2x}{5}
\]

We can also write it as:
\[
\frac{2x + 3}{5}
\]

Final answer for Problem 2:
\[
\boxed{\frac{2x + 3}{5}}
\]

---

Problem 3:
\[
\frac{3x^{-1} + x^{-2}}{4x^{-1} - x^{-2}}
\]

Remember: \(x^{-1} = \frac{1}{x}\), and \(x^{-2} = \frac{1}{x^2}\)

So rewrite the expression:

Numerator: \(\frac{3}{x} + \frac{1}{x^2}\)
Denominator: \(\frac{4}{x} - \frac{1}{x^2}\)

To combine these, get common denominators.

Numerator: \(\frac{3x}{x^2} + \frac{1}{x^2} = \frac{3x + 1}{x^2}\)
Denominator: \(\frac{4x}{x^2} - \frac{1}{x^2} = \frac{4x - 1}{x^2}\)

Now the big fraction is:
\[
\frac{\frac{3x + 1}{x^2}}{\frac{4x - 1}{x^2}}
\]

Dividing fractions → multiply by reciprocal:

\[
\frac{3x + 1}{x^2} \times \frac{x^2}{4x - 1}
\]

The \(x^2\) cancels out:

\[
\frac{3x + 1}{4x - 1}
\]

Final answer for Problem 3:
\[
\boxed{\frac{3x + 1}{4x - 1}}
\]

---

Problem 4:
\[
\frac{\frac{3}{xy} - 4}{5 - \frac{1}{4}}
\]

First, simplify the denominator: \(5 - \frac{1}{4} = \frac{20}{4} - \frac{1}{4} = \frac{19}{4}\)

Now the numerator: \(\frac{3}{xy} - 4\). Write 4 as \(\frac{4xy}{xy}\):

\[
\frac{3}{xy} - \frac{4xy}{xy} = \frac{3 - 4xy}{xy}
\]

Now the whole expression is:
\[
\frac{\frac{3 - 4xy}{xy}}{\frac{19}{4}}
\]

Divide fractions → multiply by reciprocal:

\[
\frac{3 - 4xy}{xy} \times \frac{4}{19} = \frac{4(3 - 4xy)}{19xy}
\]

Multiply out the numerator:

\[
\frac{12 - 16xy}{19xy}
\]

We can leave it like that, or factor out a negative if wanted, but it’s already simplified.

Final answer for Problem 4:
\[
\boxed{\frac{12 - 16xy}{19xy}}
\]

---

Problem 5:
\[
\frac{\frac{3}{x-1} + \frac{4}{x+2}}{\frac{7}{x+2} - \frac{5}{x-1}}
\]

This looks scary, but we’ll tackle numerator and denominator separately.

Step 1: Simplify numerator

\[
\frac{3}{x-1} + \frac{4}{x+2}
\]

Common denominator is \((x-1)(x+2)\)

So:
\[
\frac{3(x+2) + 4(x-1)}{(x-1)(x+2)} = \frac{3x + 6 + 4x - 4}{(x-1)(x+2)} = \frac{7x + 2}{(x-1)(x+2)}
\]

Step 2: Simplify denominator

\[
\frac{7}{x+2} - \frac{5}{x-1}
\]

Same common denominator: \((x-1)(x+2)\)

So:
\[
\frac{7(x-1) - 5(x+2)}{(x-1)(x+2)} = \frac{7x - 7 - 5x - 10}{(x-1)(x+2)} = \frac{2x - 17}{(x-1)(x+2)}
\]

Step 3: Put together

Now the big fraction is:
\[
\frac{\frac{7x + 2}{(x-1)(x+2)}}{\frac{2x - 17}{(x-1)(x+2)}}
\]

Divide → multiply by reciprocal:

\[
\frac{7x + 2}{(x-1)(x+2)} \times \frac{(x-1)(x+2)}{2x - 17}
\]

The \((x-1)(x+2)\) cancels out!

Left with:
\[
\frac{7x + 2}{2x - 17}
\]

Final answer for Problem 5:
\[
\boxed{\frac{7x + 2}{2x - 17}}
\]

---

Final Answer:
Problem 1: $\boxed{\frac{3}{4}}$
Problem 2: $\boxed{\frac{2x + 3}{5}}$
Problem 3: $\boxed{\frac{3x + 1}{4x - 1}}$
Problem 4: $\boxed{\frac{12 - 16xy}{19xy}}$
Problem 5: $\boxed{\frac{7x + 2}{2x - 17}}$